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I need to prove the following:

$$\sum_{n = 1}^{p - 1} n^{p - 1} \equiv (p - 1)! + p \pmod {p^2}$$

...with $p$ being an odd prime number. The statement is obviously true for$\pmod p$ because left-hand side is congruent to $-1 \pmod p$ by Fermat's little theorem, and the right-hand side also turns out to be congruent to $-1 \pmod p$ by Wilson's theorem. Now, I am not sure how to make a jump from$\pmod p$ to$\pmod {p^2}$, if that is even possible. Maybe the sum on the left could somehow be modified using the existence of the primitive root$\pmod {p^2}$.

EDIT: Elementary solution can be found here: https://mathoverflow.net/a/319824/134054

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  • $\begingroup$ @MohammadZuhairKhan Yes, thanks. I have corrected the formatting $\endgroup$ – Oldboy Dec 29 '18 at 11:56
  • $\begingroup$ This is probably true for odd $p$ $\endgroup$ – Aqua Dec 29 '18 at 12:10
  • $\begingroup$ My bad, I have corrected the statement. I apologize for that. $\endgroup$ – Oldboy Dec 29 '18 at 12:13
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    $\begingroup$ @Oldboy Isn't Fermat's little theorem, only $a^p \equiv a \pmod p$ and not for the series? $\endgroup$ – toric_actions Dec 29 '18 at 12:46
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    $\begingroup$ Elementary proof can be found here: mathoverflow.net/a/319824/134054 $\endgroup$ – Oldboy Dec 31 '18 at 17:29
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A Hint: notice that

$$\sum^{p-1}_{n=1}n^{p-1}=\frac{1}{p}\sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^n=\sum^{p}_{n=1}C^{p-n}_{p}B_{p-n}p^{n-1}$$ where $B_k$ is the $k$-th Bernoulli number.

(For further information about Bernoulli number: https://www.bernoulli.org/)

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